I want to show that if I have $\lim_{n\to\infty}a_n = a$ and $\lim_{n\to\infty}b_n = b$ then $\lim_{n\to\infty}a_nb_n = ab$

I want to do it in a slightly different way than by book but I don't know if it is correct.

$(a_n)$ converges to $a$ means $$\forall \epsilon >0, \exists n_1 \in\mathbb{N}: |a_n - a| < \frac{\epsilon}{2|b|} \,\,\,\ \forall n\in\mathbb{N}, n>n_1$$ $(b_n)$ converges to $b$ means $$\forall \epsilon >0, \exists n_2 \in\mathbb{N}: |b_n - b| < \frac{\epsilon}{2|a|} \,\,\,\ \forall n\in\mathbb{N}, n>n_2$$ Hence take $n_0 := \max\{n_1,n_2\}$ and we have $$|a_nb_n - ab| = |a_nb_n -a_nb+a_nb-ab| = |a_n(b_n-b)+b(a_n-a)| \leq |a_n|\frac{\epsilon}{2|a|}+|b|\frac{\epsilon}{2|b|}=\frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon$$

so it's proven.

Can I do that trick in the penultimate step, where I cancel out $|a_n|$ and $|a|$ ? I know that for large $n$ they will be the same, but I think this proof right now is not very rigorous!

  • 1
    $\begingroup$ For large $n$ they will not necessarily be the same, and even not necessarily be $|a_n|\leq |a|$. Take, for example $a_n=2+\frac{1}{n}$, then $a_n \to 2$ but $|a_n|>2$ for all $n$ $\endgroup$ – positrón0802 Jan 22 '17 at 19:03
  • $\begingroup$ Also, writing it like this poses a problem if $a=0$ or $b=0$. You can look at those case separately of course, or you could avoid having to distinguish cases. $\endgroup$ – StackTD Jan 22 '17 at 19:06
  • $\begingroup$ @positrón0802 yeah that is what I was worried about.. it can approach from above or below and this takes care only of the below case.. $\endgroup$ – Euler_Salter Jan 22 '17 at 19:08
  • $\begingroup$ @StackTD true, I didn't even think about that! $\endgroup$ – Euler_Salter Jan 22 '17 at 19:08
  • $\begingroup$ @StackTD I could maybe add $1$ so I have $|a|+1$ $\endgroup$ – Euler_Salter Jan 22 '17 at 19:09

Let $\epsilon > 0$. There exist $n_1 \in \mathbb{N}$ such that $$|a_n - a| < \frac{\epsilon}{2(|a|+1)} \,\,\,\ \forall n\in\mathbb{N}, n>n_1$$ There exist $n_2 \in \mathbb{N}$ such that $$|b_n - b| < \frac{\epsilon}{2(|b|+1)} \,\,\,\ \forall n\in\mathbb{N}, n>n_2$$ Also, there exists $n_3$ such that (reverse triangle inequality) $$ |a_n|-|a| \leq ||a_n|-|a||\leq |a_n-a|<1$$ for all $n > n_3$, so that $|a_n|<|a|+1$, i.e, $\frac{|a_n|}{|a|+1}<1$.

Hence, for all $n > \max\{n_1,n_2,n_3\}$ we have $$|a_nb_n - ab| = |a_nb_n -a_nb+a_nb-ab| = |a_n(b_n-b)+b(a_n-a)| \leq |a_n|\frac{\epsilon}{2(|a|+1)}+|b|\frac{\epsilon}{2(|b|+1)}<\frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon.$$


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